@harey
Hilbert's hotel tells us not to treat countably infinite sets the same way we treat finite sets. There is no 1-1 correspondence between { 1 3 5 7 9 } and { 1 2 3 4 5 6 7 8 9 10 } as there is between the (infinite set of) { odd integers } and the (infinite set of) { integers.} Hilbert's hotel always has room for more. Completion (of an infinite set of tasks) is another tricky concept. If we number some tasks 1 2 3 ... and there is no final integer, how can there be a final task? And if there's no final task how can we complete infinitely many tasks, or describe the state of things after they have transpired?
We finesse that point with a 1-1 correspondence of event times to the terms of an infinite series, 1 1/2 1/4 1/8 ... 1/2^n ...., which (conveniently) converges. We pack a countably infinite set of events into a time interval that ends at midnight. "Completion" is cleverly accomplished in two seconds. It's non-physical enough that it never could happen, but we can reasonably discuss the post-midnight state of affairs nonetheless.
We have two countably infinite sets, {coins} and {events}. At each event, two coins are added to a box and one coin from that box is discarded. So, of the two added coins, at least one is kept. We know that at midnight the entire set {coins} has entered each box, and some (proper or not) subset of them has been discarded.
The key question is this: can a coin that is kept at a certain event ever be discarded at a later event?
For Al, the answer is yes. Al always discards his lowest coin, so at each event time ti he discards coin ci. Thus eventually that is, upon completion of the infinite set of tasks, every coin that is initially kept is later discarded. At midnight no coins remain. Al's box is empty.
For Bert the answer is no. Bert discards the highest-numbered of his coins, and that is always the even coin that he just received. At no event is Bert ever scheduled to discard an odd coin. Every odd coin that enters Bert's box is kept, and it stays there forever. Bert's box contains a countably infinite set of coins.
For Charlie the answer is ... well ... um ... actually ... I guess ... yeah, but it might take an infinite number of events for it to be discarded. Well it just so happens that we have an infinite number of events that follow the keeping of every one of Charlie's initially kept coins. So, yes. All of Charlie's coins that are not immediately discarded are eventually discarded. At midnight, Charlie's box is empty.

You are shown a pile of dimes all of which have one of two distinct weights differing by a small amount not detectable by feel. Forty eight dimes are separated from this pile and you are told of these forty eight, light ones are a dime a dozen. Using a balance scale twice, find seven heavy dimes.

After N steps, they will have received 2*N coins and withdrawn N coins. At that moment, there will be 2*N-N coins in the box.
If the box is empty at midnight, this implies:
limit(2*N-N)(for N->inf) = 0
At least a little bit surprising.
@ThunderCloud
I have some troubles to refute your argument. If you remove an infinity of finite numbers from infinity of finite numbers, it does not imply no finite number remain. (Not sure I am convincing and clear enough.)
Counterargument: Al removed all coins 1 - N, coins > N remain. If N -> inf, numbering looses it's sense, but he did not remove all coins.
As for Charlie, I am ruminating, too. The first idea: every number will remain with p=1/2. Wrong, 1 will be more likely removed than 99.
2nd idea:
1st step, 2 coins: p(removing 1)=1/2
2nd step, 3 coins: p(removing 1)=p(1 was not removed in the first step) * 1/3 = 1/2 * 1/3 = 1/6, p(1 remaining after 2nd step)=1 - 1/2 - 1/6 = 1/3
3rd step, 4 coins: p(removing 1)=p(1 not yet removed) * 1/4 = 1/3 * 1/4 = 1/12, p(1 remaining after 3rd step)=1 - 1/2 - 1/6 - 1/12 =
I will not venture further, but this will not converge to 0. (Compare to 1 - 1/2 - 1/4 - 1/8...)

You have just lost your 143rd straight game of checkers and have vowed never to play another game. To confirm your vow you decide to saw your wooden checkerboard into pieces that contain no more than a single (red or black) square. With each use of the saw you may pick up a piece of the board and make one straight cut, along boundaries of individual squares. You wish to inflict as much damage as possible with each cut, so you first calculate the minimum number of saw cuts needed to finish the job. And that number is ... (spoilers appreciated.)

Gah! Mine was not a useful clue. The denominator (greater than 6 but not huge) is too large for a simulation to tell you the fraction. Also I think your simulation value is high. But you're thinking in the right direction. Here may be more useful clues.

Nice. A construction is certainly a proof of existence. A pairing without intersects exists and ... here it is!
Now I wonder if for any groups of n blue and n red points there is only one pairing without crossings?

Assuming we need to find the number in each cell, here’s a variant that will give the others a bit more information: “is the total either 9 or 21?”
this shows the others that Kleene has 4, 5, or 6. I don’t see that that’s enough yet. But maybe y’all do.